Probability Questions

Yupp · 2009-10-17 06:56:45

These are some homework questions I'm having trouble with.

1.A team is formed from selecting eight Americans and two Canadians. What is the minimum team size that guarantees at least a 90% probability that the team will not be composed of Americans only.

I'm really lost with this question.

2. Two brands of painkillers are on the market: A and B. One in four hundred people taking A suffer from side effects, and one in 1200 people taking B suffer from side effects. It is estimated that equal numbers of people take each kind of drug. If A is taken off the market, show that the probability of side effects will be halved.

I have this, but I'm doing something wrong.

P(A or B) = P(A) + P(B)
= 1/400 + 1/1200
=3/1200 + 1/200
=4/1200
=1/300
A is taken off the market.
1/1200 + 3/1200
= 4/2400
=1/600

mathsyperson · 2009-10-17 07:06:37

For 1, work out the probabilities of a team having only Americans, for various team sizes.

eg. If the team has only one member, there's an 80% chance that it will be all-American. We want it to be less than 10%, so that's not good enough.

If the team has two members, there's a 8/10 * 7/9 = 62% chance, which is still too high.
Work out the probabilities for increasing team sizes, until you get one that's low enough.

For 2, think about what would happen to 2400 people in both cases.
When both drugs are available, 1200 would take drug A and 1200 would take drug B.
3 people from the first half and 1 person from the second half would suffer side-effects, making 4 people in all.

When only B is available, all 2400 people would take that drug. Of these, 2 people would suffer side-effects, which is half the amount from before.

Yupp · 2009-10-17 07:30:45

mathsyperson wrote:

For 1, work out the probabilities of a team having only Americans, for various team sizes.
eg. If the team has only one member, there's an 80% chance that it will be all-American. We want it to be less than 10%, so that's not good enough.
If the team has two members, there's a 8/10 * 7/9 = 62% chance, which is still too high.
Work out the probabilities for increasing team sizes, until you get one that's low enough.

For 2, think about what would happen to 2400 people in both cases.
When both drugs are available, 1200 would take drug A and 1200 would take drug B.
3 people from the first half and 1 person from the second half would suffer side-effects, making 4 people in all.
When only B is available, all 2400 people would take that drug. Of these, 2 people would suffer side-effects, which is half the amount from before.

So number one would be
8/10*7/9*6/8*5/7*4/6*3/5*2/4
=403, 200/604 800
=.066666666
That's a 6% chance, therefore the team size should be 4.

Last edited by Yupp (2009-10-17 07:31:13)

mathsyperson · 2009-10-17 08:12:59

That's the right place to stop, but it's the calculation for a team of 7 people, not 4.

Yupp · 2009-10-17 08:20:03

mathsyperson wrote:

That's the right place to stop, but it's the calculation for a team of 7 people, not 4.

Alright, thanks.

Last edited by Yupp (2009-10-17 08:56:24)

bobbym · 2009-10-17 13:23:55

Hi Yupp;

What is wrong with Sorobans answer to #1 over at MHF?

Math Is Fun Forum

#1 2009-10-17 06:56:45

Probability Questions

#2 2009-10-17 07:06:37

Re: Probability Questions

#3 2009-10-17 07:30:45

Re: Probability Questions

#4 2009-10-17 08:12:59

Re: Probability Questions

#5 2009-10-17 08:20:03

Re: Probability Questions

#6 2009-10-17 13:23:55

Re: Probability Questions

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